Due Feb 24, 2:59 AM EST
How many 5-digit numbers are there that have digits 1, 2, 3, 4 and 5 (each of them exactly once)?
This is correct! We have to pick which of five digits we place on the first position, which of the remaining four digits we place on the second position and so on. We are dealing with permutations of numbers 1, 2, 3, 4 and 5 here and the answer is 5*4*3*2*1=120.
How many integer numbers between 0 and 9999 are there that have exactly one digit 7 and exactly one digit 2?
This is correct! Indeed, there are 4 ways to pick the position for digit 7. Then there are 3 positions remaining for digit 2. And then we can pick each of the remaining digits in 8 possible ways (digits 7 and 2 are forbidden). Thus by the rule of product there are 4x3x8x8=768 possible numbers.